Simplify and expand the following expression: $ \dfrac{2}{4z + 20}+ \dfrac{5}{4z + 24}- \dfrac{3z}{z^2 + 11z + 30} $
Explanation: First find a common denominator by finding the least common multiple of the denominators. Try factoring the denominators. We can factor a $4$ out of denominator in the first term: $ \dfrac{2}{4z + 20} = \dfrac{2}{4(z + 5)}$ We can factor a $4$ out of denominator in the second term: $ \dfrac{5}{4z + 24} = \dfrac{5}{4(z + 6)}$ We can factor the quadratic in the third term: $ \dfrac{3z}{z^2 + 11z + 30} = \dfrac{3z}{(z + 5)(z + 6)}$ Now we have: $ \dfrac{2}{4(z + 5)}+ \dfrac{5}{4(z + 6)}- \dfrac{3z}{(z + 5)(z + 6)} $ The least common multiple of the denominators is: $ 16(z + 5)(z + 6)$ In order to get the first term over $16(z + 5)(z + 6)$ , multiply by $\dfrac{4(z + 6)}{4(z + 6)}$ $ \dfrac{2}{4(z + 5)} \times \dfrac{4(z + 6)}{4(z + 6)} = \dfrac{8(z + 6)}{16(z + 5)(z + 6)} $ In order to get the second term over $16(z + 5)(z + 6)$ , multiply by $\dfrac{4(z + 5)}{4(z + 5)}$ $ \dfrac{5}{4(z + 6)} \times \dfrac{4(z + 5)}{4(z + 5)} = \dfrac{20(z + 5)}{16(z + 5)(z + 6)} $ In order to get the third term over $16(z + 5)(z + 6)$ , multiply by $\dfrac{16}{16}$ $ \dfrac{3z}{(z + 5)(z + 6)} \times \dfrac{16}{16} = \dfrac{48z}{16(z + 5)(z + 6)} $ Now we have: $ \dfrac{8(z + 6)}{16(z + 5)(z + 6)} + \dfrac{20(z + 5)}{16(z + 5)(z + 6)} - \dfrac{48z}{16(z + 5)(z + 6)} $ $ = \dfrac{ 8(z + 6) + 20(z + 5) - 48z} {16(z + 5)(z + 6)} $ Expand: $ = \dfrac{8z + 48 + 20z + 100 - 48z}{16z^2 + 176z + 480} $ $ = \dfrac{-20z + 148}{16z^2 + 176z + 480}$ Simplify: $ = \dfrac{-5z + 37}{4z^2 + 44z + 120}$